Testing for Uniform Continutity with Weierstrass MTest

[jbunniii]

Also, suppose the following were true:

At n=1 the original expression would be between 1/2 and 1, at n=2 it is between 1/3 and 1, and so on. THe upper bound is 1. [...] No matter what n is \frac{x}{1+nx^2} can't be more than 1.

This would NOT imply the following:

That means f_n \rightarrow 0 as n \rightarrow \infty. So that would imply uniform convergence.

In other words, $|f_n(x)| \leq 1$ for all $n$ and all $x$ does NOT imply uniform convergence, or even pointwise convergence, of $f_n$. You need $|f_n(x)| \leq B_n$ for all $n$ and all $x$, where $B_n$ is a sequence of positive numbers that converges to zero.

 

[Emspak]

OK now I get confused. You just said the maximum value for fn has to approach zero as n approaches infinity. Doesn't it? Make n big an the denominator gets way bigger than the numerator for any arbitrary n! I make n 100000000 and the original expression gets really tiny! That's approaching zero, isn't it? I am really really confused here.

 

[Office_Shredder]

You are still making the argument for pointwise convergence. As a counterexample consider if f_n=x/n was defined on all of R. For any x the sequence converges to zero because the denominator gets way bigger, but the sequence does not converge uniformly.

 

[Emspak]

Is the point here that even though fn approaches zero, and is less than some number no matter what you do with n (in this case the expression is always less than 1) that still does not imply uniform convergence?

 

[Emspak]

Dude, i am completely and utterly lost with this now.

 

[jbunniii]

Is the point here that even though fn approaches zero, and is less than some number no matter what you do with n (in this case the expression is always less than 1) that still does not imply uniform convergence?

Yes, that's right.

Here is an example: let $f_n(x)$ be the function whose graph looks like an isosceles triangle of height $1$ with one vertex at $x = 1-1/n$ and the other vertex at $x=1$, and outside of $[1-1/n, 1]$, the function is zero. This sequence satisfies $|f_n(x)| \leq 1$ for every $n$ and every $x$. The sequence converges to $0$ for every $x$, but not uniformly, because for every $n$ there is some $x$ (namely the midpoint of $[1-1/n, 1]$) with $f_n(x) = 1$.

 

[Emspak]

OK but this seems a direct contradiction to the very definition of uniform convergence above! Or is it because the maximum of fn is 1? this is why i am so confused here.

 

[jbunniii]

In other words, $f_n(x)$ looks like the attached picture.

 

[Emspak]

sorry i was referring ot the original problem, not your diagram.

 

[jbunniii]

OK but this seems a direct contradiction to the very definition of uniform convergence above! Or is it because the maximum of fn is 1? this is why i am so confused here.

You need more than just $\max(|f_n|) = 1$. You need $\max(|f_n|) = B_n$ where $B_n$ is some sequence of numbers that converges to zero as $n$ increases.

For example, $\max(|f_n|) = 1/n$ would imply uniform convergence.

 

[jbunniii]

Please check your calculation of the maximum again. Where is this function equal to zero?
$$\frac{1-nx^2}{(1+nx^2)^2}$$
Your answer should depend on $n$.

Hint: the fraction is zero whenever the numerator is zero.

 

[Emspak]

OK, now we get to choosing Bn to test for uniform convergence, yes? Would that not be, say, 1/(1+n) ? That goes to zero, doesn't it? So does -1 / (1 + n).

 

[Emspak]

It equals zero at x=1 or x=-1 when n=1. At n=4 it would have to be at x=1/2, and so on. (x would actually vary as 1/sqrt(n) to make it zero depending on what n is).

 

[jbunniii]

OK, now we get to choosing Bn to test for uniform convergence, yes? Would that not be, say, 1/(1+n) ? That goes to zero, doesn't it?

Sure, but you would have to show that your $|f_n|$ is smaller than $1/(n+1)$. It might not be, but you might be able to find some other sequence. This is why you need to calculate the maximum of $f_n$.

 

[jbunniii]

It equals zero at x=1 or x=-1 when n=1. At n=4 it would have to be at x=1/2, and so on. (x would actually vary as 1/sqrt(n) to make it zero depending on what n is).

Can you find a formula for the $x$ which gives the maximum? In other words, don't plug in $n$ values, just solve for $x$ in terms of $n$.

 

[Emspak]

when i solve for x, assuming we set the derivative expression at zero, I get:
$$\frac{1}{\sqrt{n}}= x$$. Now what?

 

[jbunniii]

when i solve for x, assuming we set the derivative expression at zero, I get:
$$\frac{1}{\sqrt{n}}= x$$. Now what?

OK, so that tells you where the maximum occurs. (Technically, you should verify that it's a maximum and not a minimum, but let's leave that aside for now.)

So now what is the value of the maximum? Plug $x = \frac{1}{\sqrt{n}}$ into $f_n(x)$ and see what you get.

 

[Emspak]

putting \sqrt{n} into the original equation, I would get $$\frac{\frac{1}{\sqrt{n}}}{1+n^2} = \frac{1}{\sqrt{n}+n^{5/2}}$$ which goes to zero the bigger n gets.

 

[jbunniii]

putting \sqrt{n} into the original equation, I would get $$\frac{\frac{1}{\sqrt{n}}}{1+n^2} = \frac{1}{\sqrt{n}+n^{5/2}}$$ which goes to zero the bigger n gets.

That's not what I get. (The $n^{5/2}$ term is wrong.)

 

[Emspak]

wait a minute. n^(1/2 ) * n^2 = n^ (1/2 + 2), no? That's 5/2.

 

[jbunniii]

Evaluate
$$f_n(x) = \frac{x}{1+nx^2}$$
at $x = \frac{1}{\sqrt{n}}$:
$$f_n\left(\frac{1}{\sqrt{n}}\right) = \frac{1/\sqrt{n}}{1 + n(1/\sqrt{n})^2} = ?$$

 

[Emspak]

Oh, SNAP! THat was stupid of me. So, you get $$\frac{1}{2\sqrt{n}}$$

 

[jbunniii]

Right. So now you have shown the following:
$$|f_n(x)| \leq \max_x |f_n(x)| = \frac{1}{2\sqrt{n}}$$
And $\frac{1}{2\sqrt{n}} \rightarrow 0$ as $n \rightarrow \infty$. What can you conclude?

 

[Emspak]

Well, that would point to uniform convergence then wouldn't it? We have a sequence of numbers that approaches zero as n--> infinity. It's also < infinity. And on top of that the absolute value of the function is less than that maximum for all n.

 

[jbunniii]

Well, that would point to uniform convergence then wouldn't it? We have a sequence of numbers that approaches zero as n--> infinity. It's also < infinity. And on top of that the absolute value of the function is less than that maximum for all n.

Yes, that's right. The conclusion is that $f_n$ converges uniformly to $0$.

 

[Emspak]

Thanks for bearing with me. This has cleared a bunch of things up, and it can get a bit frustrating.

 

[jbunniii]

No problem, this stuff is always confusing at first. Just keep doing exercises like this and it will become a lot clearer.